Question

1^2 +2^2+••••+n^2=1/6n(n+1)(2n+1)
using maths induction ​

Answer 1

Hello,

$if\ n=1\ then\ 1^2=1\ and\ \dfrac{1}{6}*1*2*3=1:\ true\ for\ n=1$

We suppose the property true for n:

1²+2²+...+n²=n(n+1)(2n+1) / 6

and we are going to demonstrate that the property is true for n+1:

1²+2²+..+(n+1)²=(n+1)*(n+2)*(2n+3)/6

$1^2+2^2+...+n^2+(n+1)^2\\\\=n*(n+1)*(2n+1)/6+(n+1)^2\\\\=(n+1)/6*[n(2n+1)+6n+6]\\\\=(n+1)/6*(2n^2+7n+6)\\\\=(n+1)(n+2)(2n+3)/6$